Mixing
gradient of quadratic form
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I am new to linear algebra. I appreciate if somebody helps me to solve this problem. I have a function $f(v)=hvv^Hh^H$, where $hin mathbf{C}^{1times N}$ is a constant vector, $vin mathbf{C}^{Ntimes 1}$ is variable vector, and $H$ denotes Hermitian transpose. I need to calculate $nabla f(v)$
matrices derivatives
asked Dec 19 '18 at 17:51
zstrzstr
11
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add a comment |
$begingroup$
I am new to linear algebra. I appreciate if somebody helps me to solve this problem. I have a function $f(v)=hvv^Hh^H$, where $hin mathbf{C}^{1times N}$ is a constant vector, $vin mathbf{C}^{Ntimes 1}$ is variable vector, and $H$ denotes Hermitian transpose. I need to calculate $nabla f(v)$
matrices derivatives
asked Dec 19 '18 at 17:51
zstrzstr
11
$endgroup$
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Btw, this resource is quite useful for these sorts of queries math.uwaterloo.ca/~hwolkowi/matrixcookbook.pdf
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– ted
Jan 9 at 22:04
add a comment |
$begingroup$
I am new to linear algebra. I appreciate if somebody helps me to solve this problem. I have a function $f(v)=hvv^Hh^H$, where $hin mathbf{C}^{1times N}$ is a constant vector, $vin mathbf{C}^{Ntimes 1}$ is variable vector, and $H$ denotes Hermitian transpose. I need to calculate $nabla f(v)$
matrices derivatives
asked Dec 19 '18 at 17:51
zstrzstr
11
$endgroup$
I am new to linear algebra. I appreciate if somebody helps me to solve this problem. I have a function $f(v)=hvv^Hh^H$, where $hin mathbf{C}^{1times N}$ is a constant vector, $vin mathbf{C}^{Ntimes 1}$ is variable vector, and $H$ denotes Hermitian transpose. I need to calculate $nabla f(v)$
matrices derivatives
matrices derivatives
asked Dec 19 '18 at 17:51
zstrzstr
11
asked Dec 19 '18 at 17:51
zstrzstr
11
asked Dec 19 '18 at 17:51
zstrzstr
11
asked Dec 19 '18 at 17:51
zstrzstr
11
asked Dec 19 '18 at 17:51
zstrzstr
11
11
$begingroup$
Btw, this resource is quite useful for these sorts of queries math.uwaterloo.ca/~hwolkowi/matrixcookbook.pdf
$endgroup$
– ted
Jan 9 at 22:04
add a comment |
$begingroup$
Btw, this resource is quite useful for these sorts of queries math.uwaterloo.ca/~hwolkowi/matrixcookbook.pdf
$endgroup$
– ted
Jan 9 at 22:04
$begingroup$
Btw, this resource is quite useful for these sorts of queries math.uwaterloo.ca/~hwolkowi/matrixcookbook.pdf
$endgroup$
– ted
Jan 9 at 22:04
$begingroup$
Btw, this resource is quite useful for these sorts of queries math.uwaterloo.ca/~hwolkowi/matrixcookbook.pdf
$endgroup$
– ted
Jan 9 at 22:04
add a comment |
1 Answer
1
active
oldest
votes
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For consistency, use column vectors so that both ${,h,vinmathbb C}^{Ntimes 1}$
Then consider the complex scalar
$$phi = h^Tv$$
in terms of which your (real) function is
$$f= |phi|^2 = phi^*phi$$
Since $phi$ depends solely upon $v$ (and $phi^*$ upon $v^*$) we can easily find the gradient as
$$eqalign{
df &= phi^*dphi = (phi^*h)^Tdv cr
frac{partial f}{partial v} &= phi^*h cr
}$$
Taking the complex conjugate yields
$$
frac{partial f}{partial v^*} = phi h^*
implies
frac{partial f}{partial v^H} = phi h^H
$$
answered Dec 22 '18 at 4:50
greggreg
8,0451822
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Thanks for your help Greg.
$endgroup$
– zstr
Dec 31 '18 at 19:36
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For consistency, use column vectors so that both ${,h,vinmathbb C}^{Ntimes 1}$
Then consider the complex scalar
$$phi = h^Tv$$
in terms of which your (real) function is
$$f= |phi|^2 = phi^*phi$$
Since $phi$ depends solely upon $v$ (and $phi^*$ upon $v^*$) we can easily find the gradient as
$$eqalign{
df &= phi^*dphi = (phi^*h)^Tdv cr
frac{partial f}{partial v} &= phi^*h cr
}$$
Taking the complex conjugate yields
$$
frac{partial f}{partial v^*} = phi h^*
implies
frac{partial f}{partial v^H} = phi h^H
$$
answered Dec 22 '18 at 4:50
greggreg
8,0451822
$endgroup$
$begingroup$
Thanks for your help Greg.
$endgroup$
– zstr
Dec 31 '18 at 19:36
add a comment |
$begingroup$
For consistency, use column vectors so that both ${,h,vinmathbb C}^{Ntimes 1}$
Then consider the complex scalar
$$phi = h^Tv$$
in terms of which your (real) function is
$$f= |phi|^2 = phi^*phi$$
Since $phi$ depends solely upon $v$ (and $phi^*$ upon $v^*$) we can easily find the gradient as
$$eqalign{
df &= phi^*dphi = (phi^*h)^Tdv cr
frac{partial f}{partial v} &= phi^*h cr
}$$
Taking the complex conjugate yields
$$
frac{partial f}{partial v^*} = phi h^*
implies
frac{partial f}{partial v^H} = phi h^H
$$
answered Dec 22 '18 at 4:50
greggreg
8,0451822
$endgroup$
$begingroup$
Thanks for your help Greg.
$endgroup$
– zstr
Dec 31 '18 at 19:36
add a comment |
$begingroup$
For consistency, use column vectors so that both ${,h,vinmathbb C}^{Ntimes 1}$
Then consider the complex scalar
$$phi = h^Tv$$
in terms of which your (real) function is
$$f= |phi|^2 = phi^*phi$$
Since $phi$ depends solely upon $v$ (and $phi^*$ upon $v^*$) we can easily find the gradient as
$$eqalign{
df &= phi^*dphi = (phi^*h)^Tdv cr
frac{partial f}{partial v} &= phi^*h cr
}$$
Taking the complex conjugate yields
$$
frac{partial f}{partial v^*} = phi h^*
implies
frac{partial f}{partial v^H} = phi h^H
$$
answered Dec 22 '18 at 4:50
greggreg
8,0451822
$endgroup$
For consistency, use column vectors so that both ${,h,vinmathbb C}^{Ntimes 1}$
Then consider the complex scalar
$$phi = h^Tv$$
in terms of which your (real) function is
$$f= |phi|^2 = phi^*phi$$
Since $phi$ depends solely upon $v$ (and $phi^*$ upon $v^*$) we can easily find the gradient as
$$eqalign{
df &= phi^*dphi = (phi^*h)^Tdv cr
frac{partial f}{partial v} &= phi^*h cr
}$$
Taking the complex conjugate yields
$$
frac{partial f}{partial v^*} = phi h^*
implies
frac{partial f}{partial v^H} = phi h^H
$$
answered Dec 22 '18 at 4:50
greggreg
8,0451822
answered Dec 22 '18 at 4:50
greggreg
8,0451822
answered Dec 22 '18 at 4:50
greggreg
8,0451822
answered Dec 22 '18 at 4:50
greggreg
8,0451822
8,0451822
$begingroup$
Thanks for your help Greg.
$endgroup$
– zstr
Dec 31 '18 at 19:36
add a comment |
$begingroup$
Thanks for your help Greg.
$endgroup$
– zstr
Dec 31 '18 at 19:36
$begingroup$
Thanks for your help Greg.
$endgroup$
– zstr
Dec 31 '18 at 19:36
$begingroup$
Thanks for your help Greg.
$endgroup$
– zstr
Dec 31 '18 at 19:36
add a comment |
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$begingroup$
Btw, this resource is quite useful for these sorts of queries math.uwaterloo.ca/~hwolkowi/matrixcookbook.pdf
$endgroup$
– ted
Jan 9 at 22:04